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Distance Minimization

  1. Sep 29, 2007 #1
    Find the point P on the parabola y = x^2 closest to the point (1,0).

    To solve for P...

    I used the distance formula, and then took the derivate. Using points (x, x^2) and (1,0)...

    however, while taking the derivative I get a really nasty 4th power equation, which I am not sure I can solve without a calculator.
  2. jcsd
  3. Sep 29, 2007 #2


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    Can you derive your solution here?
  4. Sep 29, 2007 #3
    I don't know how you did it, but the big hint here is of course to minimize the distance *squared*, rather than the distance.
  5. Sep 30, 2007 #4
    ok this is what I have so far...

    D = sqrt. ((x-2)^2 + (x^2)^2)

    then I took the derivative...

    dD/dx = (1/2)((x-1)^2 + x^4)^-1/2)*(2(x-1) + 4x^3)

    this simplifies to..

    dD/dx = ((x-1) +2x^3)/sqrt.((x-1)^2 + x^4)

    setting that equal to zero...

    this is where I encounter issues..

    2x^3 + x - 1 =0


    sqrt. ((x-1)^2 + x^4) = 0
  6. Sep 30, 2007 #5


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    As Dead Wolfe suggested, try taking the derivative of D2 instead.
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